What are dependent variables?

agasfer [191]

Answer:

Dunes are loose sand hills.

Loess is a compacted wind-blown formation of sediment.

Explanation:

Sand dunes are, as their name implies, made of sand which is itself made of tiny quartz pieces. Sand dunes are present where there is a ready source of broken down sandstone or other type of rock and wind to transport the sand. The dunes are mobile and loose and migrate over time.

Loess comes in vast formations and was created in a very different way. In the last ice age, winds from the north carried sediments loosened and ground by glaciers southwards. There, the sediments accumulated in large formations and became compacted over time.

6 0

3 years ago

A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out

scZoUnD [109]

Answer: The mass percent of potassium bromide in the mixture is 9.996%

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For lead (II) bromide:

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0

3 years ago

Ce cantitate de O se gaseste in 5,8 g hidroxid de magneziu

coldgirl [10]

Answer:

separare pâlnii de picurare balon cotat. 1. 2. 3. 4 5 6 7. 8. 9. 1 ... Mod de lucru: 25 g Na2SO4∙7H2O se dizolvă în cantitatea minim ... Exemple: NaOH – hidroxid de sodiu.

Explanation:

make me as brain liest

4 0

3 years ago

2. Why are there so many volcanoes in California?

Kruka [31]

Answer:

Most higher risk volcanoes are far from California’s largest cities and several produce heat that’s used to generate electricity in what are the world’s most productive geothermal power plants, such as the Salton Buttes, 160 miles southeast of Los Angeles, and the Clear Lake Volcanic Field 85 miles north of San Francisco that powers the Geysers steam field.

Explanation:

5 0

3 years ago

Liquid A and liquid B form a solution that behaves ideally according to Raoult's law. The vapor pressures of the pure substances

Rama09 [41]

Answer:

Vapor pressure of solution → 151.1 Torr

Option 2.

Explanation:

Raoult's Law is relationed to colligative property about vapor pressure. A determined solute, can make, the vapor pressure of solution decreases.

ΔP = P° . Xm

where Xm is the mole fraction of solute, P° (vapor pressure of pure solvent)

and ΔP = Vapor pressure of pure solvent - Vapor pressure of solution.

In order to determine the vapor pressure of solution, we need to determine, the vapor pressure of B and A in the solution

B's pressure = P° B . Xm

When we add A to B, A works as the solute and B, as the solvent.

Vapor pressure of pure B is 135 torr. (P° B)

In order to determine, the Xm, we use the moles of A and B

Xm = 5.3 mol of B / (1.28 + 5.3) → 0.806

B's pressure = 135 Torr . 0.806 → 108.81 Torr

If mole fraction of B is 0.806, mole fraction for A (solute) will be (1 - 0.806)

A's pressure = 218 Torr . 0.194 → 42.3 Torr

Vapor pressure of solution is sum of vapor pressures of solute + solvent.

Vapor pressure of solution = 42.3 Torr + 108.81 Torr → 151.1 Torr

6 0

2 years ago